Let G=Sym(n) .Let C be a cyclic group. Suppose that then how to show that the generators for in G all have the same cycle type and so are conjugate in G.
I am assuming that the cyclic group is finite.
If generates then generates where is relatively prime with the order of because a homomorphic image of a cyclic group is cyclic. Let for disjoint cycles . The order of is the least common multiple of the lengths of all these cycles. Therefore, divides for all . Now if is an integer so that then certainly . Therefore, if is an -cycle then still stays as an -cycle. Thus, we have where is same cycle length as which means and have the same cycle type and consequently they are conjugate.